A physical quantity C is related to four other quantities p, q, r and s as follows $ C = \frac{pq^2}{r^3 \sqrt{s}} $ The percentage errors in the measurement of p, q, r and s are 1%, 2%, 3% and 2% respectively. The percentage error in the measurement of C will be _______ %.
To find the percentage error in the measurement of C, we first need to understand how errors propagate through the given formula. The expression for C is:
C = \(\frac{pq^2}{r^3 \sqrt{s}}\)
The percentage error formula for a product or quotient involving powers, given a function like \(Z = \frac{A^m B^n}{C^p D^q}\), is:
\(\frac{\Delta Z}{Z} \times 100 \approx m \frac{\Delta A}{A} \times 100 + n \frac{\Delta B}{B} \times 100 + p \frac{\Delta C}{C} \times 100 + q \frac{\Delta D}{D} \times 100\)
Applying this to our expression:
The percentage error in C is:
\(\frac{\Delta C}{C} \times 100 \approx 1 \cdot 1\% + 2 \cdot 2\% + 3 \cdot 3\% + 0.5 \cdot 2\%\)
Calculating this gives:
Add these errors together:
1% + 4% + 9% + 1% = 15%
Thus, the percentage error in the measurement of C is 15%, which is within the expected range of 15,15.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)